WORLD FOR CIVIL

Solution - Exam Problem 2

The structure shown below is a truss which is pinned to the floor at point A and supported by a roller at point H. For this structure:
A. Draw a Free Body Diagram showing all support forces and loads.
B. Determine the value of all the support forces acting on the structure.
C. Determine the force in member DG by method of sections.
Unless otherwise indicated, all joints and support points are assumed to be pinned or hinged joints.

Solution:

STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces, as well as any needed angles and dimensions.

STEP 2: Break any forces not already in x and y direction into their x and y components.
STEP 3: Apply the equilibrium conditions.
Sum F_x = A_x = 0
Sum F_y = A_y + H_y -12,000 lbs - 20,000 lbs - 10,000 lbs = 0
Sum T_A = (-12,000 lbs)(20 ft) - (20,000 lbs)(40 ft) - (10,000 lbs)(60 ft) + H_y(80 ft) = 0
Solving for the unknowns: A_y = 21,500 lbs; H_y = 20,500 lbs

PART C - Now find the internal force in member DB by section method. Cut vertically through members DF, DG, and EG. Analyze right hand section.

STEP 1: Draw a free body diagram of the right hand section (see diagram).

STEP 2: Resolve all forces into x and y components (see diagram).

STEP 3: Apply equilibrium conditions:
Sum F_x = EG + DG cos (51.3^o) + DF cos (22.6^o) = 0
Sum F_y = -10,000 lbs + 20,500 lbs - DG sin (51.3^o) - DF sin (26.6^o) = 0
Sum T_G = -DF cos (26.6^o)(15 ft) + (20,500 lbs)(20 ft) = 0
Solving for the unknowns: DF = 30,600 lbs (C); DG = -4,090 (opposite direction) or DG= 4,090 lbs (T); EG = -24,800(opposite direction) or EG=24,800 lbs (T)